It depends on the scales which you are interested in. During the deSitter evolution of the universe (let's assume exact deSitter and completely flat potential for simlicity) the fluctuations of the inflaton field exit the horizon and freeze in with a power spectrum of $H_*^2/(2\pi)^2$ where $*$ denotes that the value of the Hubble parameter is taken at the time of horizon crossing ($k = aH$). The power spectrum of the curvature perturbation ($A_s^2$) on superhorizon scales is then
$$
\begin{equation}
{\cal{P}}_\zeta \simeq \frac{H_*^4}{(2\pi\dot\phi_*)^2}
\end{equation}
$$
Since the curvature perturbation is conserved on superhorizon scales (in the absence of isocurvature perturbations) as long as the scales you are interested in stay outside the horizon this should give the correct amplitude. So far this is all standard text book stuff.
Now in your scenario with two inflationary phases, if the scales of interest never re-enter the horizon then you get the scalar amplitude as above because the curvature perturbation is conserved on super-horizon scales.
However, if the scale of interest first exits the horizon during the first inflationary phase then re-enters the horizon during the intermediary phase and then exits again during the second inflationary phase then the amplitude will be changed because perturbations evolve inside the horizon. How exactly they will evolve will depend on the dynamics during the phase between the two inflationary phases.
Then you should track how your fluctuations evolve. For example, in the flat gauge the fluctuations of the inflaton evolve according to the following equation:
$$
\ddot{\delta\phi} + 3H\dot{\delta\phi} + \left[\frac{k^2}{a^2}+V''(\phi)-\frac{8\pi G}{a^3}\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{a^3}{H}\dot\phi^2\right)\right]\delta\phi=0
$$